In class, we have shown that F(N) forms a real vector space

1. l^p-spaces

Let F (N) be the space of all sequence of real numbers. In class, we have shown that F(N) forms a real vector space. Let 1 ≤ p < ∞ and let l^p(N) be the subset of F(N) consisting of sequences a = such that P∞

n=1|a(n)|^p < ∞.

Lemma 1.1. The subset l^p(N) forms a vector subspace of F(N).

Proof. It is easy to see that 0 ∈ l^p(N). Let a ∈ l^p(N) and k ∈ R. By the property of convergence of infinite series,P∞

n=1|ka(n)|^p = |k|^pP∞

n=1|a(n)|^p < ∞. Hence ka ∈ l^p(N). If a, b ∈ l^p(N), by Minkowski inequality for infinite series, we have

n

X

n=1

|a(n) + b(n)|^p

!1/p

≤

n

X

n=1

|a(n)|^p

!1/p

+

n

X

n=1

|b(n)|^p

!1/p

< ∞.

Hence a + b ∈ l^p(N). 

For each a ∈ l^p(N), we set

kak_p =

∞

X

n=1

|a(n)|^p

!1/p

.

Lemma 1.2. Let (xn) be a sequence of real numbers such that xn≤ M for any n ≥ 1. If (x_n) is convergent to x in R, then x ≤ M.

Proof. Exercise. 

Theorem 1.1. l^p(N) with k · kp is a Banach space over R.

Proof. By the property of convergence of infinite series, kkak_p= |k|kak_p. (This is also shown in the previous lemma). For any a, b ∈ l^p(N), the Minkowski inequality for infinite series is exactly the triangle inequality: ka + bk_p ≤ kak_p+ kbk_p. By definition, kak_p ≥ 0. Now let us prove that kak_p = 0 if and only if a = 0. For each n ∈ N and each m ∈ N with m ≥ n, we have

|a(n)|^p ≤

m

X

n=1

|a(n)|^p ≤

∞

X

n=1

|a(n)|^p= kak^p_p. Hence for each i ∈ N, one has

(1.1) |a(n)| ≤ kak_p.

Hence kak_p = 0, then |a(n)| = 0 for any n ∈ N and hence a(n) = 0 for any n ∈ N. This shows that a = 0 by definition. If a = 0, it is obvious that kak_p = 0. We prove that k · k_p defines a norm on l^p(N).

Let us prove that l^p(N) is complete with respect to the norm k · kp. Let (a_i)^∞_i=1 be a Cauchy sequence in l^p(N). For any  > 0 there exists N ∈ N such that kai− a_jk_p < /2 whenever i, j ≥ N_. By (1.1), we have

|a_i(n) − aj(n)| ≤ kai− a_jk_p<  2

whenever i, j ≥ N_. This shows that the sequence (a_i(n))^∞_i=1 of real numbers is a Cauchy sequence in R for any n ∈ N. Let us denote a(n) to be the limit of (ai(n))^∞_i=1, i.e.

i→∞lim ai(n) = a(n).

Hence we obtain a sequence of real numbers a : N → R. Let us show that a ∈ l^p(N).

1

2

Since (ai)^∞_i=1 is a Cauchy sequence in l^p(N), it is bounded in l^p(N). There exists M > 0 such that

ka_ik_p ≤ M for any i ≥ 1.

For any m ≥ 1, we have

m

X

n=1

|a_i(n)|^p≤

∞

X

n=1

|a_i(n)|^p = ka_ik^p_p≤ M^p.

Since (ai(n))^∞_i=1 is convergent in R for each n ∈ N, by the property of limit,

i→∞lim

m

X

n=1

|a_i(n)|^p =

m

X

n=1

|a(n)|^p. By Lemma 1.2, P_m

n=1|a(n)|^p ≤ M^p for any m ≥ 1. By Lemma 1.2 again, (consider x_m = Pm

n=1|a(n)|^p for each m ≥ 1.)

∞

X

n=1

|a(n)|^p ≤ M^p < ∞.

This implies that a ∈ l^p(N). Now let us prove that (ai)^∞_i=1 is convergent to a in l^p(N). For each m ≥ 1 and any i, j ≥ N,

m

X

n=1

|a_i(n) − aj(m)|^p≤ ka_i− a_jk^p_p <

 2

p

. For any m ≥ 1, and any i, j ≥ N, Pm

n=1|a_i(n) − aj(n)|^p< (/2)^p. By Lemma 1.2, for any i ≥ N_,

m

X

n=1

|a_i(n) − a(n)|^p = lim

j→∞

m

X

n=1

|a_i(n) − a_j(n)|^p≤ 2

p

. Therefore for each m ≥ 1, and for i ≥ N_, we have

m

X

n=1

|a_i(n) − a(n)|^p≤ 2

p

. By Lemma 1.2 again, (taking m → ∞,), for i ≥ N,

∞

X

n=1

|a_i(n) − a(n)|^p = lim

m→∞

m

X

n=1

|a_i(n) − a(n)|^p ≤ 2

p

. This implies that kai− ak ≤ /2 <  for i ≥ N_. We prove that limi→∞ai = a.